All categories

2 years ago

Rachel works at a bookstore. On Tuesday, she sold twice as many books as she did on Monday. On Wednesday, she

Mathematics

1 answer:

ICE Princess25 [194]2 years ago

6 0

Answer:

m +2m +(2m -6) = 19

Step-by-step explanation:

m = number of books sold on Monday.

2m = number of books sold on Tuesday, twice as many as Monday.

2m -6 = number of books sold on Wednesday, 6 fewer than Tuesday.

The total number of books sold on the three days gives the equation you're looking for:

m +2m +(2m -6) = 19

_____

<em>Additional comment</em>

Solving this gives ...

5m = 25 . . . . add 6

m = 5 . . . . . divide by 5

5, 10, and 4 books were sold on the three days.

You might be interested in

Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.

const2013 [10]

Answer:

$T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

Step-by-step explanation:

Remember that for any curve $r(t)$

The tangent vector is given by

$T(t) = \frac{r'(t) }{| r'(t)| }$

And the normal vector is given by

$N(t) = \frac{T'(t)}{|T'(t)|}$

For this case, using the chain rule

$r'(t) = ( -10*2tsin(t^2) , 102t cos(t^2) )\\$

And also remember that

$|r'(t)| = \sqrt{(-10*2tsin(t^2))^2 + ( 10*2t cos(t^2) )^2} \\\\ = \sqrt{400 t^2*( sin(t^2)^2 + cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t$

Therefore

$T(t) = r'(t) / |r'(t) | = ( -10*2tsin(t^2) , 10*2t cos(t^2) )/ 20t\\\\ = ( -10*2tsin(t^2)/ 20t , 10*2t cos(t^2) / 20t )\\= ( -sin(t^2), cos(t^2) )$

Similarly, using the quotient rule and the chain rule

$T'(t) = ( -2t cos(t^2) , -2t sin(t^2))$

And also

$|T'(t)| = \sqrt{ ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t$

Therefore

$N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

Notice that

1. $|N(t)| = |T(t) | = \sqrt{ cos(t^2)^2 + sin(t^2)^2 } = \sqrt{1} = 1$

2. $N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0$

Simlarly

$r'(t) = (2t,-6,0) \\$

and

$|r'(t)| = \sqrt{(2t)^2 + 6^2} = \sqrt{4t^2 + 36}$

Therefore

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

Then

$T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)$

and also

$|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 + ( (3 t)/(9 + t^2)^{3/2})^2 + 0^2 }\\= 3/(t^2 + 9 )$

And since

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

6 0

2 years ago

A class of 12 girls and 14 boys is going to choose 3 students to represent them in student government. Allie suggested that the

Vikki [24]

<h2>Answer:</h2>

The method suggested by Bobby is better than Allie.

<h2>Step-by-step explanation:</h2>

A fair sample is a sample where each sample has an equal opportunity to get selected for any survey.

Here, in the given question, Allie suggested that the first 3 students to arrive in class the next day should be the representatives. This sample is not random or fair.

Whereas, Bobby suggested that each student roll 2 numbers cubes and the 3 students with the highest sum should be the representatives. This method is a fair method.

The method suggested by Bobby is better than Allie as every student will get an equal chance to perform and also has an equal chance to win and get a high sum.

8 0

3 years ago

Read 2 more answers

Lim (sin(pi/x)) as x tends to 0 from both sides

strojnjashka [21]

Hello from MrBillDoesMath!

Answer:

Limit does not exist.

Discussion:

The function 1/x has a vertical asymptote as x approaches 0 so

sin (pi/x) has no limit as x approaches 0. In fact, it oscillates wildly between -1 and 1 as x approaches 0. See attached graph of function

Thank you,